#23214 [Bgs->Opn]: When exactly is mysql_insert_id() optional?
| From: | philip@php.net | Date: | Sun, 20 Apr 2003 15:53:18 +0000 |
| Subject: | #23214 [Bgs->Opn]: When exactly is mysql_insert_id() optional? | ||
| References: | 1 | Groups: | php.doc |
| Request: | Send a blank email to phpdoc+get-969352743@lists.php.net to get a copy of this message | ||
ID: 23214
Updated by: philip@php.net
-Summary: mysql_insert_id - Never use the function "Blank"
Reported By: Stefan_255 at hotmail dot com
-Status: Bogus
+Status: Open
Bug Type: Documentation problem
Operating System: win2k
PHP Version: 4.3.0
New Comment:
Please explain what the disclaimers are, we shouldn't rely on user
comments as part of the manual.
Stefan, some questions: Do you have multiple database connections? Can
you create a short short that demonstrates the problem? (one that
connects/selects to mysql..). This also might just be some old bug
that was fixed as 4.3.0 works. If it was an introduced feature (i.e.
resource was made optional) we need to document when that happened.
Reopening...
Previous Comments:
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[2003-04-20 09:58:45] momo@php.net
there is enough disclaimers on the site comments.
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[2003-04-14 17:20:55] Stefan_255 at hotmail dot com
I copied a few lines of code from the net and got it to work in my own
test environment. When I placed the software on the production server
(linux PHP 4.1.2) - it only "almost" worked.
The problem was that it had lines with statements like:
mysql_insert_id();
where strange id's sometimes were returned !
Then, when I changed the code to:
mysql_insert_id($con); // The resource link_identifier
everything was fine. I believe that what happened was that my program
received id's from somebody else doing INSERT on the same server?
My conclusion was:
Never use the function without the
"resource link_identifier" !
And then again, - Why should you?
In the documentation it says:"If link_identifier isn't specified, the
last opened link is assumed."
My code was simply:
$sql= "INSERT INTO xrefs (referrer) VALUES ('$ref')";
echo "sql $sql <br>";
$result = mysql_query($sql,$con);
//$referrer_id = mysql_insert_id(); // FAILED
$referrer_id = mysql_insert_id($con); // WORKED OK
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Edit this bug report at http://bugs.php.net/?id=23214&edit=1